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%% lecture16.tex
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%% Made by Alex Nelson
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%% Started on  Fri Dec 10 12:52:01 2010 Alex Nelson
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If we have
\begin{equation}
\varphi\colon\mathscr{G}\to\mathfrak{gl}(V)
\end{equation}
a representation, and $\mathscr{H}\propersubset\mathscr{G}$ is
the Cartan subalgebra, then we recall a weight vector $v$ is an
eigenvector
\begin{equation}
\varphi(h)v = \lambda(h)v
\end{equation}
for every $h\in\mathscr{H}$. We have an adjoint representation
\begin{equation}
\ad\colon\mathscr{G}\to\mathfrak{gl}(\mathscr{G})
\end{equation}
where
\begin{equation}
(\ad x)v = [x,v]
\end{equation}
and the weight vectors for this representation are called root
vectors, and the weights are called roots. To define roots and
root vectors for $\mathscr{G}$ we are solving
\begin{equation}
[h,v] = \alpha(h)v.
\end{equation}
We can construct new root vectors from a given root vector by
applying $e_{i}$, $f_{j}$ to it. We have
\begin{subequations}
\begin{align}
[h,e] = \alpha(h)e & \iff he=eh+\alpha(h)e \\
 & \iff h(ev) = \big(eh+\alpha(h)e\big)v
\end{align}
\end{subequations}
but 
\begin{equation}
hv = \lambda(h)v \quad\implies\quad h(ev) =
\big(\lambda(h)+\alpha(h)\big) ev.
\end{equation}
This is either zero or another distinct eigenvector.

\begin{prop}
If $v$ is a weight vector with weight $\lambda$ and $e$ is a root
vector with root $\alpha$, then $ev$ is a weight vector with
weight $\lambda+\alpha$ unless $ev=0$.
\end{prop}

Now we will introduce a definition. Well several
definitions. First we introduce a notion of a Cartan Matrix which
is presented differently in different papers.
\begin{defn}
A \define{Cartan Matrix} is a matrix $A=[a_{ij}]$ such that
\begin{enumerate}
\item $a_{ii}=2$ are the diagonal components;
\item $a_{ij}\in\ZZ$ for any $i$, $j$;
\item $a_{ij}\leq 0$ for off-diagonal components;
\item although not necessarily symmetric, if $a_{ij}=0$ then
  $a_{ji}=0$;
\item it should be symmetrizable, i.e. we have a diagonal matrix
  $B$ such that $AB=D$ is also diagonal.
\end{enumerate}
\end{defn}
\begin{rmk}
Most of the time we will work with $A$ nondegenerate, i.e.
\begin{equation}
\det(A)\not=0
\end{equation}
But this is not a necessary condition, so we do not make it part
of the definition.
\end{rmk}
For every classical Lie Algebra, the matrix $a_{ij}$ % from the homework assignment,
is nondegenerate
\begin{equation}
\det(a_{ij})\not=0.
\end{equation}
We have explicitly computed this, so we should look at our
answers and nothing more.

The only thing that needs discussion is ``Why is $A$
symmetrizable?'' We know there exists a nondegenerate invariant
inner product on classical Lie algebras. The adjoint
representation is orthogonal with respect to this inner product,
i.e.\
\begin{equation}
\big\<[h,x], y\big\>+\big\<x,[h,y]\big\> = 0
\end{equation}
where $h,x,y\in\mathscr{G}$. This could be viewed as a
consequence of compact Lie groups having unitary representations
giving invariant inner product.

We can introduce the Killing form\index{Killing Form!Invariant Inner Product, Relation to} as
\begin{equation}
\<x,y\> = \tr\big(\ad_{x}\ad_{y}\big)
\end{equation}
which is an invariant inner product. %
%% %% Incorrect proof given on the day of the lecture
%% We consider the case when
%% $h\in\mathscr{H}$, $x=e_{j}$, and $y=f_{k}$, then we see that
%% \begin{subequations}
%% \begin{align}
%% \big\<[h,x], y\big\>+\big\<x,[h,y]\big\> &= \big\<[h,e_{j}],
%% f_{k}\big\>+\big\<e_{j},[h,f_{k}]\big\> =\\
%% &= \big\<a_{ij}e_{j}, f_{k}\big\>+\big\<e_{j},-a_{ik}f_{k}]\big\>
%% \end{align}
%% \end{subequations}
%% \dots
%% we should get the formula
%% \begin{equation}
%% a_{ij}\<e_{i},f_{i}\> = a_{ji}\<e_{j}, f_{j}\>
%% \end{equation}
%% %% Correct proof given in the next lecture
We may introduce an invariant inner product on the group
\begin{equation}
\<Ux,Uy\> = \<x,y\>
\end{equation}
where $U\in G$, but when $U=\1+u\varepsilon$ where $\varepsilon$
is ``small'', then we get
\begin{equation}
\<\varphi(u)x,y\> + \<x,\varphi(u)y\> = 0
\end{equation}
where $\varphi$ is a morphism. But as a representation we have
\begin{equation}
\Big\<[u,x],y\Big\> + \Big\<x,[u,y]\Big\> = 0.
\end{equation}
If we let $u=e_{i}$, $x=f_{i}$, $y=h_{j}$ we get
\begin{subequations}
\begin{align}
\Big\<[u,x],y\Big\> + \Big\<x,[u,y]\Big\>
&= \Big\<[e_{i},f_{i}],h_{j}\Big\> + \Big\<f_{i},[e_{i},h_{j}]\Big\>\\
&= \<h_{i},h_{j}\> + \<f_{i}, -a_{ji}e_{i}\>
\end{align}
\end{subequations}
This holds if and only if
\begin{equation}
\begin{diagram}[small,hug,height=13.5pt]
\<h_{i}, h_{j}\> & \rEq & a_{ji}\<f_{i},e_{i}\> \\
\dEq             &      & \dEq \\
\<h_{j}, h_{i}\> & \rEq & a_{ij}\<f_{j},e_{i}\>
\end{diagram}
\end{equation}
Using the inner product on the group, we may construct  the
matrix $B=\diag\<e_{i},f_{i}\>$ which implies $AB$ is symmetric.

For every Cartan matrix, we may construct a Lie algebra called a
Kac--Moody algebra. Really %irreducible
simple Lie Algebras are Kac--Moody algebras with additional
condition that the Cartan matrix is positive definite. We will
now describe all irreducible representations of classical Lie
Algebras; this is true for all Lie Algebras related to compact
groups, and reductive Lie Algebras.

In reality we may say for every compact group, the corresponding
Lie algebras have precisely the right generators. Moreover, we
may classify algebras of compact groups. This gives us a general
theorem for representations of compact Lie algebras. We would
like to explain the notion of a highest weight vector in this
situation. Namely the highest weight vector $v$ is such that
\begin{equation}
\varphi(e_{i})v = 0
\end{equation}
for all $e_{i}\in\mathscr{G}$. Of course this means that
\begin{equation}
\varphi(h)v = \lambda(h)v
\end{equation}
for all $h\in\mathscr{H}$, then this $\lambda$ is called the
highest weight. 

First of all, what is $\lambda(-)$? It is a linear functional
$\lambda\in\mathscr{H}^{*}$, i.e.
\begin{equation}
\lambda\colon\mathscr{H}\to\FF
\end{equation}
it is a linear functional acting on the Cartan subalgebra. Then:
\begin{enumerate}
\item Irreducible representations contain not more than one
  highest weight vector, up to a constant factor; 
\item Finite dimensional irreducible representation $\iff$ finite
  dimensional representation with one highest weight vector;
\item For every $\lambda\in\mathscr{H}^{*}$ one can construct a
  unique irreducible representation with highest weight $\lambda$
  but this representation can be infinite dimensional;
\item\label{lec16:cartan:mostImportantPoint}\marginpar{\eqref{lec16:cartan:mostImportantPoint} is the most important point!}
  This representation is finite dimensional if and only if
  $\lambda(h_{i})$ is a non-negative integer.
\end{enumerate}
This gives us a complete description of finite dimensional
irreducible representations.

\subsection{Dynkin Diagrams}
It is a very convenient way to depict Cartan matrices. Namely %it is 
first of all the dimension of the Cartan algebra is called the
\define{Rank of the Lie Algebra}. We have $\ClassicalGroup{A}_{\ell}$,
$\ClassicalGroup{B}_{\ell}$, $\ClassicalGroup{C}_{\ell}$,
$\ClassicalGroup{D}_{\ell}$ all of rank $\ell$. 

\begin{wrapfigure}[2]{r}{4cm}
  \vspace{-20pt}
  \begin{center}
    \includegraphics{img/LieImg.3}
  \end{center}
  \vspace{-20pt}
\end{wrapfigure}
The Dynkin diagram for $\ClassicalGroup{A}_{\ell}$ we draw $\ell$
vertices and we draw edges. We have the number of edges connecting vertices
$v_{i}$ to $v_{j}$ be given by the formula using the Cartan matrix 
\begin{equation}
n_{ij}=a_{ij}a_{ji}.
\end{equation}
Suppose we know $B$, its diagonal so we only need to keep track
of one index really. Since we suppose we know $B$, then
\begin{equation}
a_{ij}b_{j}=a_{ji}b_{i}
\end{equation}
implies
\begin{equation}
b_{i} = \frac{a_{ij}b_{j}}{a_{ji}}
\end{equation}
we get
\begin{equation}
n_{i}b_{i} = {a_{ij}}^{2}b_{j},
\end{equation}
or equivalently
\begin{equation}
{a_{ij}}^{2} = \frac{n_{i}b_{i}}{b_{j}}.
\end{equation}
What is the conclusion? If we know $[n_{i}]$ and $[b_{i}]$, we can
compute $[a_{ij}]$. 

Let us write down the Dynkin diagrams for the classical Lie
groups we have considered.
\begin{wrapfigure}[4]{r}{4.5cm}
  \vspace{-20pt}
  \begin{center}
    \includegraphics{img/LieImg.4}
  \end{center}
  \vspace{-20pt}
\end{wrapfigure}
\noindent{}For $\ClassicalGroup{D}_{n}$ we have the diagram drawn
on the right for the case when $n=7$ (observe there are 7
vertices). The Cartan matrix for $\ClassicalGroup{D}_{n}$ is
symmetric. One can observe this by considering the adjacency
matrix for the graph.

\begin{wrapfigure}[2]{l}{4.5cm}
  \vspace{-20pt}
  \begin{center}
    \includegraphics{img/LieImg.5}
  \end{center}
  \vspace{-20pt}
\end{wrapfigure}
\noindent{}For $\ClassicalGroup{C}_{n}$ we see the Cartan
matrix is not symmetric, but we can symmetrize it. We find that
$a_{n-1,n}a_{n,n-1}=2$.

\begin{wrapfigure}[2]{r}{4.5cm}
  \vspace{-20pt}
  \begin{center}
    \includegraphics{img/LieImg.6}
  \end{center}
  \vspace{-20pt}
\end{wrapfigure}
\noindent{}For $\ClassicalGroup{B}_{n}$ we see the Dynkin diagram
is ``the same'' as for $\ClassicalGroup{C}_{n}$ but with
different labels for the vertices.

Almost all of these groups are simple and almost all of them are
not isomorphic. But almost all. For example, in
$\ClassicalGroup{D}_{2}$ we have two disconnected vertices for
the Dynkin diagram. So $\ClassicalGroup{D}_{2}$ is not simple, it
is the direct product of $\SU{2}$ at the level of Lie Algebras,
and \emph{almost} the direct product at the level of Lie
groups. So we may examine the Dynkin diagram for $\ClassicalGroup{D}_{3}$
to find:
\begin{center}
\includegraphics{img/LieImg.7}
\end{center}
\noindent{}So we see this is the same Dynkin diagram as for
$\ClassicalGroup{A}_{3}$ which implies at the level of Lie
Algebras
\begin{equation}
\SU{4}\iso\SO{6}
\end{equation}
but only at the level of Lie Algebras. We similarly have $\ClassicalGroup{B}_{2}\iso\ClassicalGroup{C}_{2}$
by inspection of the Dynkin diagrams, but again it is an
isomorphism at the level of Lie Algebras.

\input{tex/box1}

\subsection{Returning to Representations}
The representations are described by means of highest weight. We
had previously
\begin{equation}
\varphi(e_{i})x = 0
\end{equation}
where $x$ is our highest weight vector, and the highest weight is
described by
\begin{equation}
\varphi(h)x = \lambda(h)x
\end{equation}
where $\lambda\in\mathscr{H}^{*}$ is the highest weight. We
should demand $\lambda(h_{i})\geq0$, and
$\lambda(h_{i})\in\ZZ$. We will now turn our attention to examples.

We will consider the fundamental representations of
$\ClassicalGroup{A}_{\ell+1}=\mathfrak{sl}(\ell+1)$. The
fundamental representation is the representation by
$(1+\ell)\times(1+\ell)$ matrices. We found
\begin{equation}
e_{i}=E_{i,i+1}
\end{equation}
where $E_{i,j}$ has zero components everywhere except at $i,j$ it
is 1. The Cartan subalgebra is
\begin{equation}
\mathscr{H}=\left\{\begin{bmatrix}\lambda_1 & & \\
 & \ddots & \\
 &        & \lambda_{n}
\end{bmatrix}\text{ such that }\lambda_{1}+\dots+\lambda_{\ell+1}=0\right\}
\end{equation}
What are the weight vectors here? It is quite clear that the
weight vectors $u_{1}$, \dots, $u_{\ell+1}$ are the standard
basis vectors. Observe
\begin{equation}
hu_{i}=\lambda_{i}u_{i}.
\end{equation}
What is the highest weight vector of this representation? We see
that
\begin{equation}
e_{i}u_{j} = 0
\end{equation}
unless $j=i+1$ we have
\begin{equation}
e_{i}u_{i+1}=u_{i}
\end{equation}
The highest weight vector is clearly $u_{1}$ because the shift
goes down and there is no way down. This implies the
representation is irreducible as the highest weight vector is
unique up to some coefficient. We see that
\begin{equation}
\lambda(h_{i}) = \delta_{i1}
\end{equation}
also holds.

What about the tensor product of representations. We find the
basis to be $u_{j}\otimes u_{k}$ and
\begin{subequations}
\begin{align}
h_{i}(u_{j}\otimes u_{k}) &= (h_{i}u_{j})\otimes u_{k} + u_{j}\otimes(h_{i}u_{k})\\
&= (\lambda_{j}+\lambda_{k})(u_{j}\otimes u_{k})
\end{align}
\end{subequations}
We found all the weight vectors\dots well not really since
$u_{1}\otimes u_{2}$ is a weight vector with the same weight as
$u_{2}\otimes u_{1}$, so $u_{1}\otimes u_{2}\pm u_{2}\otimes
u_{1}$ is again a weight vector. We find the highest weight
vector to be $u_{1}\otimes u_{1}$, but we see that
\begin{equation}
e_{1}(u_{1}\otimes u_{2}) = u_{1}\otimes u_{1}
\end{equation}
and
\begin{equation}
e_{1}(u_{2}\otimes u_{1}) = u_{1}\otimes u_{1}
\end{equation}
so it follows that $u_{1}\otimes u_{2}-u_{2}\otimes u_{1}$ is
again a highest weight vector\dots so we have 2 distinct highest
weight vectors! This cannot be an irreducible
representation. This we know, we may consider the symmetric and
antisymmetric parts of the representation.



